Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)

\(=x^2y^2+a^2b^2+x^2b^2+a^2y^2\)

\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)

\(=\left(b^2+y^2\right)\left(x^2+a^2\right)\)

\(m^2-16+n^2-2mn\)

\(=n^2-2mn+m^2-16\)

\(=\left(n-m\right)^2-16\)

\(=\left(n-m-4\right)\left(n-m+4\right)\)

m² - 16 + n² - 2mn

= m² - 2mn + n² - 16

= (m - n)² - 4²

= (m - n - 4)(m - n + 4)

m² -16 +n² - 2mn = (m² - 2mn + n²) -16

=( m - n )² - 4²

= (m- n- 4)(m - n + 4)

Ta có: \(m^2+m+2=m^2+2m-m+2=m\left(m+2\right)-\left(m+2\right)=\left(m+2\right)\left(m-1\right)\)

a: \(x^3z+x^2yz-x^2z^2-xyz^2\)

\(=x^2z\left(x+y\right)-xz^2\left(x+y\right)\)

\(=xz\left(x+y\right)\left(x-z\right)\)

a: Ta có: \(m^2+2mn+n^2-p^2+2pq+q^2\)

\(=\left(m+n\right)^2-\left(p-q\right)^2\)

\(=\left(m+n-p+q\right)\left(m+n+p-q\right)\)

Lời giải:

$N=p^{m+2}q-pq^{m+3}-p^{m+3}q^{n+4}$

$=pq(p^{m+1}-q^{m+2}-p^{m+2}q^{n+3})$

(x^m+2)+(x^m) = 2x^m+2 = 2(x^m+1)

(x^x+1)-(x^x)-1 = x^x+1-x^x-1 = 0

(m^4)-(n^4) = (m²)²-(n²)² = (m²-n²)(m²+n²)

\(=m^2-\left(n-2\right)^2=\left(m-n+2\right)\left(m+n-2\right)\)